PR 4 # 30

Wacky · 2006-07-13, 21:09

If the hour and minute hands of a watch are interchanged, how many different possible times could the watch show?

retina · 2006-07-14, 11:01

Infinitely many times can be displayed, it all depends on how much precision you want to measure with.

Wacky · 2006-07-14, 11:27

If the hands are in their "normal" configuration, then an infinite number of times could be represented. However, when you swap the hands, many times cannot be properly represented.

For example, midnight can be represented in either configuration because at that time both hands are straight up. However, the time one minute later cannot be represented with the hands swapped.

retina · 2006-07-14, 23:21

With the hands swapped you can represent times on other worlds :) You never stated they had to be valid times for people using the 12 hour system that live on Earth. BTW: I have a watch that has 24 hours as one rotation for the hour hand. Can I use that watch?

Wacky · 2006-07-15, 00:54

Sorry, retina.

I am unwilling to accept "alternative" interpretations until some someone comes up with a solution that meets the "obvious" conditions, or shows that they are hopelessly ambiguous.

R. Gerbicz · 2006-07-15, 12:35

Suppose that the correct time is A hours and x minutes, the interchanged time is B hours and y minutes. Here 0<=A,B<12 and integers, 0<=x,y<60 and real numbers, because these are times.

Using degrees the hour hand at 30*A+x/2 position and the minute hand at 6*x position. So the required equations are:
30*A+x/2=6*y
6*x=30*B+y/2

From the first equation: y=5*A+x/12
Using this and the second equation: 6*x=30*B+5*A/2+x/24 so
143/24*x=30*B+5*A/2, from this x=60/143*(A+12*B)
Suppose that A and B fix then we can determine x by the previous formula and 0<=x is true and x<=60/143*(11+12*11)=60 so A and B can't be both 11, but in other cases 0<=x<60 is true.
0<=y=5*A+x/12<5*11+60/12=60 what is needed. So the number of the solutions is 12*12-1=143.

2006-07-13, 21:09	#1
Wacky Jun 2003 The Texas Hill Country 3²·11² Posts	PR 4 # 30 If the hour and minute hands of a watch are interchanged, how many different possible times could the watch show?

2006-07-14, 11:01	#2
retina Undefined "The unspeakable one" Jun 2006 My evil lair 1100001010010₂ Posts	Infinitely many times can be displayed, it all depends on how much precision you want to measure with.

2006-07-14, 11:27	#3
Wacky Jun 2003 The Texas Hill Country 3²·11² Posts	If the hands are in their "normal" configuration, then an infinite number of times could be represented. However, when you swap the hands, many times cannot be properly represented. For example, midnight can be represented in either configuration because at that time both hands are straight up. However, the time one minute later cannot be represented with the hands swapped.

2006-07-14, 23:21	#4
retina Undefined "The unspeakable one" Jun 2006 My evil lair 2×11×283 Posts	With the hands swapped you can represent times on other worlds :) You never stated they had to be valid times for people using the 12 hour system that live on Earth. BTW: I have a watch that has 24 hours as one rotation for the hour hand. Can I use that watch?

2006-07-15, 00:54	#5
Wacky Jun 2003 The Texas Hill Country 3²·11² Posts	Sorry, retina. I am unwilling to accept "alternative" interpretations until some someone comes up with a solution that meets the "obvious" conditions, or shows that they are hopelessly ambiguous.

2006-07-15, 12:35	#6
R. Gerbicz "Robert Gerbicz" Oct 2005 Hungary 10111001110₂ Posts	Suppose that the correct time is A hours and x minutes, the interchanged time is B hours and y minutes. Here 0<=A,B<12 and integers, 0<=x,y<60 and real numbers, because these are times. Using degrees the hour hand at 30A+x/2 position and the minute hand at 6x position. So the required equations are: 30A+x/2=6y 6x=30B+y/2 From the first equation: y=5A+x/12 Using this and the second equation: 6x=30B+5A/2+x/24 so 143/24x=30B+5A/2, from this x=60/143(A+12B) Suppose that A and B fix then we can determine x by the previous formula and 0<=x is true and x<=60/143(11+1211)=60 so A and B can't be both 11, but in other cases 0<=x<60 is true. 0<=y=5A+x/12<511+60/12=60 what is needed. So the number of the solutions is 1212-1=143.